Hence, we can conclude that the absolutemaximum value of f(x) in the interval [0, 2π] is 2πoccurring at x = 2π and the absolute minimum value of f(x)in the interval [0, 2π] is 0 occurring at x = 0.

Question - 13 : -
Find two numbers whose sum is 24 and whose product is aslarge as possible.

Answer - 13 : -

Let one number be x. Then, theother number is (24 − x).

Let P(x) denote theproduct of the two numbers. Thus, we have:

∴By second derivative test, x = 12 is thepoint of local maxima of P. Hence, the product of the numbers isthe maximum when the numbers are 12 and 24 − 12 = 12.

Question - 14 : -
Find two positive numbers x and y suchthat x + y = 60 and xy³ ismaximum.

Answer - 14 : -

The two numbers are x and y suchthat x + y = 60.

⇒ y = 60 − x

Let f(x) = xy^3.

∴By second derivative test, x = 15 is apoint of local maxima of f. Thus, function xy³ ismaximum when x = 15 and y = 60 − 15 = 45.

Hence, the required numbers are 15 and 45.

Question - 15 : -
Find two positive numbers x and y suchthat their sum is 35 and the product x²y⁵ is a maximum

Answer - 15 : -

Let one number be x. Then, theother number is y = (35 − x).

Let P(x) = x²y⁵. Then, we have:

x =0, x = 35, x = 10

When x =35,

and y = 35 − 35 = 0.This will make the product x² y⁵ equalto 0.

When x = 0, y =35 − 0 = 35 and the product x²y² willbe 0.

∴ x = 0 and x = 35 cannotbe the possible values of x.

When x = 10, we have:

∴ By second derivative test, P(x) willbe the maximum when x = 10 and y = 35 − 10 =25.

Hence, the required numbers are 10 and 25.

Question - 16 : -
Find two positive numbers whose sum is 16 and the sum ofwhose cubes is minimum.

Answer - 16 : -

Let one number be x. Then, theother number is (16 − x).

Let the sum of the cubes of these numbers bedenoted by S(x). Then,

Now,

∴ By second derivative test, x = 8 is thepoint of local minima of S.

Hence, the sum of the cubes of the numbers is the minimumwhen the numbers are 8 and 16 − 8 = 8.

Question - 17 : -
A square piece of tin of side 18 cm is to made into a boxwithout top, by cutting a square from each corner and folding up the flaps toform the box. What should be the side of the square to be cut off so that thevolume of the box is the maximum possible?

Answer - 17 : -

Let the side of the square to be cut offbe x cm. Then, the length and the breadth of the box will be(18 − 2x) cm each and the height of the box is x cm.

Therefore, the volume V(x)of the box is given by,

V(x)= x(18 − 2x)²

x = 9or x = 3

If x = 9, then the lengthand the breadth will become 0.

x ≠ 9.

x = 3.

Now,

By second derivative test, x =3 is the point of maxima of V.

Hence, if we remove a square of side 3 cm from each cornerof the square tin and make a box from the remaining sheet, then the volume ofthe box obtained is the largest possible.

Question - 18 : -
A rectangular sheet of tin 45 cm by 24 cm is to be madeinto a box without top, by cutting off square from each corner and folding upthe flaps. What should be the side of the square to be cut off so that thevolume of the box is the maximum possible?

Answer - 18 : -

Let the side of the square to be cut offbe x cm. Then, the height of the box is x, thelength is 45 − 2x, and the breadth is 24 − 2x.

Therefore, the volume V(x)of the box is given by,

Now,.

x = 18and x = 5

It is not possible to cut off a square ofside 18 cm from each corner of the rectangular sheet. Thus, x cannotbe equal to 18.

∴x = 5

Now,

By second derivative test, x =5 is the point of maxima.

Hence, the side of the square to be cut off to make thevolume of the box maximum possible is 5 cm.

Question - 19 : -
Show that of all the rectangles inscribed in a given fixedcircle, the square has the maximum area.

Answer - 19 : -

Let a rectangle of length l andbreadth b be inscribed in the given circle of radius a.

Then, the diagonal passes through the centre and is oflength 2a cm.

Now, by applying the Pythagoras theorem, we have:

∴Area of the rectangle,

By the second derivativetest, when, then the area of the rectangle is the maximum.

Since, the rectangle is a square.

Hence, it has been proved that of all the rectanglesinscribed in the given fixed circle, the square has the maximum area.

Question - 20 : -
Show that the right circular cylinder of given surface andmaximum volume is such that is heights is equal to the diameter of the base.

Answer - 20 : -

Let r and h bethe radius and height of the cylinder respectively.

Then, the surface area (S) ofthe cylinder is given by,

Let V be the volume of thecylinder. Then,

∴ By second derivative test, the volume is the maximumwhen

Hence, the volume is the maximum when the height is twicethe radius i.e., when the height is equal to the diameter.

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